An introduction to Wave diffraction & the

Wiener–Hopf technique

Ian Thompson

Department of Mathematical Sciences

What is the Wiener–Hopf technique?

• A method for solving problems involving discontinuous boundaryconditions.

• Uses classical complex analysis:

• poles & residues,

• branch points,

• Cauchy & Fourier integrals,

• Liouville’s theorem,

• Jordan’s lemma.

• Surrounded by an aura of fear, thanks largely to some terribleliterature.

1

The Sommerfeld problem

x

y

barrier

shadow region

reflection region

Θ

Incident waves

• There is a fundamental physics problem here, as well as a difficultmaths problem.

2

The Sommerfeld problem (continued)

• Arnold Sommerfeld solved his famous problem in 18961.

• His method doesn’t seem to work for any other problems!

• Around 1931, rigourous work on integral equations by N. Wiener,E. Hopf and others 2 was adapted to solve the Sommerfeld problem3.

• This came to be known as the Wiener–Hopf technique.

• It enables us to solve many otherwise intractable problems.

• Later, D. S. Jones found an easier way to apply the method usingFourier integrals.

• This strips away a lot of unnecessary mathematical apparatus.1Sommerfeld: ‘Optics’ (1964) [In English]2Paley & Wiener: ‘Fourier transforms in the complex domain’ (1934)3Wiener: ‘Collected Works’ 1981 [In English]

3

How it works

• Let Γ be a contour that divides C into upper and lower regions.

• Represent the wavefield φ in the form

φ(x, y) =1

2πi

∫

Γ

φ̂(α; y)e−iαx dα.

• Suppose that φ̂(α; 0) → 0 as |α| → ∞.

• Jordan’s lemma says that for x > 0, we can close Γ in the lowerhalf plane, whereas for x < 0 we can close in the upper half plane.

• This is absolutely crucial:

Different singularity structures above and below Γ givedifferent results for x > 0 and x < 0.

4

The wave equation

• We have (

∇2 − 1

c2

∂2

∂t2

)

Φ(x, y, t) = 0.

• Assume that fields are time-harmonic , i.e.

Φ(x, y, t) = ℜ[φ(x, y)e−iωt

].

• Now we have to solve the Helmholtz equation

(∇2 + k2)φ(x, y) = 0, k = ω/c,

• Note propagation directions, eg. eiky propagates upwards, because

φ = eiky ⇒ Φ = ℜ[

ei(ky−ωt)]

.

5

Setting up the Sommerfeld problem

• Write φt = φi + φs.

• The incident wave is known:

φi = eik(x cos Θ+y sin Θ).

• We need to solve the Helmholtzequation subject to the b.c.

φs(x, 0) = −eikx cos Θ on x > 0.

• Also, ∇φs ∼ O(r−1/2) as

r =√

x2 + y2 → 0 (tip condition).

• φs must not include any contributionsthat are incoming toward the barrier, orthat grow as |y| is increased.

x

y

φt = 0

Θ

Incident waves

6

The first golden rule

• Always look to exploit symmetries.

• Decompose the incident wave into odd and even (in y) functions:

φiS(x, y) =

1

2

[φi(x, y) + φi(x,−y)

], φi

A(x, y) =1

2

[φi(x, y) − φi(x,−y)

].

• φiA(x, 0) = 0, so this satisfies the b.c. on its own.

• In other words, φsA(x, y) ≡ 0, and so φs(x, y) = φs

S(x, y).

• φs(x, 0) must be continuous on x < 0 (where there is no barrier),and it’s an even function, so

∂φs

∂y= 0 y = 0, x < 0.

7

The second golden rule

• Exhaust all elementary arguments before resorting to W–H.

• We need to solve for φs(x, y) in y ≥ 0.

• Start by writing

φs(x, y) =1

2πi

∫

Γ

φ̂(α; y)e−iαx dα.

• Apply operator (∇2 + k2) (ok to commute with integral):∫

Γ

[

∂2φ̂

∂y2− (α2 − k2)

]

e−iαx dα = 0.

• This can only be true ∀x if [·] ≡ 0, and hence

φ̂(α; y) = A(α)e−γ(α)y + B(α)eγ(α)y,

where γ(α) = (α2 − k2)1/2.

8

The functions A and B

φs(x, y) =1

2πi

∫

Γ

[

A(α)e−γ(α)y + B(α)eγ(α)y]

e−iαx dα.

• At points on Γ where ℜ[γ(α)] = ℜ[(α2 − k2)1/2] 6= 0 :

If e−γ(α)y → 0 as y → ∞, then |eγ(α)y| → ∞ & vice-versa.

• At points on Γ where ℜ[γ(α)] = ℜ[(α2 − k2)1/2] = 0 :

If e−γ(α)y represents an outgoing contribution then eγ(α)y

represents an incoming contribution & vice-versa.

• One solution is always unphysical—doesn’t matter which becausewe have yet to choose the branch of γ(α).

• It is conventional to take B(α) ≡ 0.

• Also, A(α) ∼ O(α−η) with η > 1 as α → ∞ ∈ Γ.

9

The contour Γ

φs(x, y) =1

2πi

∫

Γ

A(α)e−γ(α)ye−iαx dα.

• Recall: Γ divides C into upper and lower regions.

• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .

• γ(α) =√

α2 − k2 for real α with |α| > k.

• Go clockwise around α = −k to reduce theargument from 0 to −π/2.

• Go anticlockwise around = k to change itback to 0.

• D±= {α ± iu : α ∈ Γ, u ≥ 0}.

−k

k ℜ[α]

ℑ[α]

10

The contour Γ

φs(x, y) =1

2πi

∫

Γ

A(α)e−γ(α)ye−iαx dα.

• Recall: Γ divides C into upper and lower regions.

• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .

• γ(α) =√

α2 − k2 for real α with |α| > k.

• Go clockwise around α = −k to reduce theargument from 0 to −π/2.

• Go anticlockwise around = k to change itback to 0.

• D±= {α ± iu : α ∈ Γ, u ≥ 0}.

−k

k ℜ[α]

ℑ[α]

10

The contour Γ

φs(x, y) =1

2πi

∫

Γ

A(α)e−γ(α)ye−iαx dα.

• Recall: Γ divides C into upper and lower regions.

• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .

• γ(α) =√

α2 − k2 for real α with |α| > k.

• Go clockwise around α = −k to reduce theargument from 0 to −π/2.

• Go anticlockwise around α = k to change itback to 0.

• D±= {α ± iu : α ∈ Γ, u ≥ 0}.

−k

k ℜ[α]

ℑ[α]

10

The contour Γ

φs(x, y) =1

2πi

∫

Γ

A(α)e−γ(α)ye−iαx dα.

• Recall: Γ divides C into upper and lower regions.

• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .

• γ(α) =√

α2 − k2 for real α with |α| > k.

• Go clockwise around α = −k to reduce theargument from 0 to −π/2.

• Go anticlockwise around α = k to change itback to 0.

• D±= {α ± iu : α ∈ Γ, u ≥ 0}.

−k

k ℜ[α]

ℑ[α]

10

The contour Γ

φs(x, y) =1

2πi

∫

Γ

A(α)e−γ(α)ye−iαx dα.

• Recall: Γ divides C into upper and lower regions.

• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .

• γ(α) =√

α2 − k2 for real α with |α| > k.

• Go clockwise around α = −k to reduce theargument from 0 to −π/2.

• Go anticlockwise around α = k to change itback to 0.

• D±= {α ± iu : α ∈ Γ, u ≥ 0}.

−k

k ℜ[α]

ℑ[α]

D+

D−

10

Boundary Condition on x > 0

φs(x, y) =1

2πi

∫

Γ

A(α)e−iαx dα.

• We need to satisfy φs(x, 0) = −eikx cos Θ on x > 0.

• So, if x > 0, the only contribution must come from a simple poleat α = −k cos Θ, otherwise we get the wrong x dependence.

• Γ passes above α = −k cos Θ, and we can write

A(α) =1

α + k cos Θ+ A

−(α),

• A−(α) is unknown, but it is analytic in D−

, and A−(α) → 0 as

α → ∞ ∈ D−.

• Jordan’s lemma applies (close Γ in D−); the BC on x > 0 is

satisfied.

11

Symmetry Condition on x < 0

∂φs

∂y(x, 0) =

−1

2πi

∫

Γ

G(α)e−iαx dα where G(α) = γ(α)A(α).

• We need to satisfy∂φs

∂y(x, 0) = 0 on x < 0.

• If x < 0, there must be no contributions to the integral.

• Hence, G(α) = G+(α), i.e. it’s is analytic in D+

, and G+(α) → 0

as α → ∞ ∈ D+.

• Jordan’s lemma applies again (close Γ in D+); the symmetry

condition on x > 0 is now satisfied.

12

Wiener–Hopf equation

• We have G(α) = γ(α)A(α), and hence

G+(α) = γ(α)

[1

α + k cos Θ+ A−(α)

]

. (∗)

• We can product split γ(α) = (α2 − k2)1/2:

γ+(α) = (α + k)1/2; γ

−(α) = (α − k)1/2.

• Use in (∗): G+(α)

γ+(α)=

γ−(α)

α + k cos Θ+ γ

−(α)A−(α).

• LHS is analytic in D+. RHS is analytic in D− \ {−k cos Θ}.

• Subtractγ−(−k cos Θ)

α + k cos Θfrom both sides. . .

13

Wiener–Hopf equation (continued)

G+(α)

γ+(α)− γ

−(−k cos Θ)

α + k cos Θ︸ ︷︷ ︸

=γ−(α) − γ

−(−k cos Θ)

α + k cos Θ+ γ

−(α)A−(α)

︸ ︷︷ ︸

.

Analytic in D+Analytic in D−

• Both sides represent the same function! Call it J(α).

• This is analytic in D+ ∪ D−= C, i.e. it’s entire.

• We need to determine its behaviour as |α| → ∞.

• From the LHS, J(α) → 0 as α → ∞ ∈ D+.

• It’s not clear what happens to γ−(α)A

−(α) as α → ∞ ∈ D−

.

14

Tauberian Theorems

• Suppose that

f(x) =1

2πi

∫

Γ

F−(α)e−iαx dα,

where F−(α) → 0 as α → ∞ ∈ D−

, so that f(x) = 0 for x > 0.

• If f(x) ∼ O(xη) as x → 0−

(with η > −1) then F−(α) ∼

O(α−(η+1)) as α → ∞ ∈ D−

• We have to apply this to φt, because φt(x, 0) = 0 for x > 0.

φt(x, 0) =1

2πi

∫

Γ

⌣

[1

α + k cos Θ+ A

−(α)

]

e−iαx dα

(the residue at α = −k cos Θ is eikx cos Θ = φi(x, 0)).

• ∇φt ∼ O(r−1/2) as r → 0, so φt(x, 0) ∼ O(1) as x → 0−

(at

most). =⇒ A−(α) ∼ O(α−1) as α → ∞ ∈ D−

(at most).

15

Solving the Wiener–Hopf equation

• At last, we have everything we need. . .

G+(α)

γ+(α)− γ

−(−k cos Θ)

α + k cos Θ︸ ︷︷ ︸

=γ−(α) − γ

−(−k cos Θ)

α + k cos Θ+ γ

−(α)A−(α)

︸ ︷︷ ︸

= J(α).

→ 0 as α → ∞ ∈ D+ → 0 as α → ∞ ∈ D−

• J(α) → 0 as α → ∞ in any direction, and it’s an entire function.

• Liouville’s theorem says that J(α) ≡ 0.

• Now we can complete the solution using either G+

or A−:

G+(α) = γ(α)A(α) =⇒ A(α) =

γ−(−k cos Θ)

γ−(α)(α + k cos Θ).

16

What does it all mean?

φs(x, y) =1

2πi

∫

Γ

γ−(−k cos Θ)

γ−(α)(α + k cos Θ)e−γ(α)|y|e−iαx dα.

• This satisfies all of the required conditions.

• In this (very special) case, the integral can be evaluated exactly(!)

φs(r, θ) =eikr

2

{

w[

eiπ/4√

2kr sin(

θ+Θ2

)]

+ w[

eiπ/4√

2kr sin(

θ−Θ2

)]}

.

• w(z) is the scaled complex error function; w(z) = e−z2[1− erf(z)].

• This is Sommerfeld’s original solution.

• The error function rapidly but continuously (de)activates the planewave terms as the optical boundaries are crossed.

• There is also a circular wave that is O[(kr)−1/2] in the far field,away from the boundaries.

17

k = 1.0, Θ = π/4

-

−20−20

20

202.52.5

(Plane wave terms included)

17

k = 1.0, Θ = π/4

-

−20−20

20

202.52.5

(Plane wave terms excluded—diffraction only)

17

k = 1.0, Θ = π/4

-

−20−20

20

202.52.5

(Plane wave terms excluded—diffraction only)

17

One more thing. . .

• To solve the Sommerfeld problem, we had to write

γ(α) = γ+(α)γ

−(α).

• This is called a kernel factorisation.

• It was easy to find γ±(α); just ‘share out’ the singularities:

γ(α) = (α2 − k2)1/2, γ±(α) = (α ± k)1/2.

• It doesn’t take much to make this hard, eg.

f(α) = 1 + γ(α), f±(α) =?

18

Cauchy integrals

• Suppose that K(α) → 1 as α → ∞ ∈ Γ. Define

I+(α) =

1

2πi

∫

Γ

⌣log[K(z)]

z − αdz, α ∈ D+

,

I−(α) =

−1

2πi

∫

Γ

⌢log[K(z)]

z − αdz, α ∈ D−

.

• Then K±(α) = exp[I

±(α)], because

K+(α)K

−(α) = exp[I+(α) + I

−(α)] = exp[log[K(α)]].

• We use functions with known factorisations to make K(α) → 1:

f(α) = 1 + γ(α), K(α) =1 + γ(α)

γ(α), f

±(α) = γ

±(α)K

±(α).

• It is usually very hard to evaluate these integrals, even numerically.

19

References

• The Sommerfeld problem appears in lots of books, eg.

• Linton & McIver (2001) ‘Handbook of Mathematical techniques for wave/structure interactions’,

• Billingham & King (2000) ‘Wave motion’,

• Crighton et al. (1992) ‘Modern Methods in Analytical Acoustics’.

• Jones (1964) ‘The Theory of Electromagnetism’,

• Linton & McIver solve one more problem via the Wiener–Hopfmethod.

• Only the dreaded

• Noble (1988) ‘Methods based on the Wiener–Hopf technique’.

seems to contain further examples.

• Don’t read this after dark when there is a full moon.

• Prof. David Abrahams is in the process of writing a new book. . .

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